UNSTABLE NEUTRON
By Prof. L. Kaliambos (Natural Philosopher in New Energy) July 24 , 2015 Gell-mann in 1964 proposed that a nucleon consists of three charged quarks like the up and down having the fractional charges +2e/3 and - e/3 respectively. So they are able to give charge distributions in nucleons responsible for revealing the strong electromagnetic forces under the applications of natural lows. However in 1973 influenced by the fallacious meson theory(1935) and by Einstein’s invalid massless quanta of fields (EXPERIMENTS REJECT RELATIVITY) he abandoned the well-established electromagnetic laws in favor of invalid ideas and introduced the hypothesis of strange “color forces” between false massless gluons in his theory of quantum chromodynamics. In fact, massles particles cannot exist in accordance with my DISCOVERY OF PHOTON MASS. Under this physics crisis in my paper “Nuclear structure is governed by the fundamental laws of electromagnetism” (2003) I showed that the so-called strong interaction is due to a strong electromagnetic interaction between the 9 extra charged quarks in protons and the 12 extra charged quarks in neutrons which led to my discovery of 288 quarks in nucleons.( See my NEW STRUCTURE OF PROTONS AND NEUTRONS). The extra charged quarks among the 288 quarks exert strong electromagnetic forces of short range in a strong electromagnetic interaction of short range. (See my DISCOVERY OF NUCLEAR FORCE AND STRUCTURE ). Surprisingly applications of electromagnetic laws on such charge distributions which give for proton extra (4u,5d) charged quarks and for the neutron extra (8d,4u) charged quarks lead exactly to the simplest nuclear binding (-2.2246 MeV) of the deuterium. Moreover such extra quark led to the discovery of 288 quarks in nucleons. As a result the proton has 93 (dud) neutral quark triads. Among them there are 4u charged quarks distributed along the periphery and 5d charged quarks limited in the center. Whereas the neutron has 92 (dud) neutral quark triads and among them are distributed 8d charged quarks along the periphery and 4u charged quarks limited in the center So, the structure of protons and neutrons is given by PROTON = [ 93(dud) + 4u +5d ] = 288 quarks NEUTRON = [ 92(dud) + 8d + 4u]= 288 quarks After a careful analysis we found the masses of down (Md) and of up quark (Mu) as Md = 3.69348645 MeV/c2 and Mu = 2.40016645 MeV/c2 which give not only the masses Mn = 939.565378 MeV/ c2 and Mp = 938.272046 MeV/c2 of neutron and proton respectively but also the difference Mn - Mp = 1.293332 MeV/c2 which is exactly equal to Md - Mu. That is, Md - Mu = 3.69348645 - 2.40016645 = 1.293332 MeV. However in "Down quark- WIKIPEDIA”''' and in "Up quark-WIKIPEDIA" one can see the following confusing values as Md = ( 4.1 - 5.7) = 4.9 MeV or a so-called precise value Md = 4.79 MeV. Mu = (1.7 - 3.1) = 2.4 MeV or a so-called precise value Mu = 2.01 MeV.' Of course these values are wrong because the difference Md - Mu = 4.79 - 2.01 = 2.78 MeV/c2 is greater than the correct value Mn - Mp = 1.293332 MeV. '''WHY THE FREE NEUTRON IS AN UNSTABLE PARTICLE' So far under the fallacious theories of the WRONG STANDARD MODEL no one was able to explain why the free neutron is an unstable particle. For example the simple quark model of (udd) scheme for neutron and (uud) scheme for proton cannot explain what happens inside a nucleon with the fallacious massless gluons and why the neutron is an unstable particle. In fact, using the absorption of the antineutrino by the correct structure of proton one sees that the stable structure of proton turns into the unstable structure of neutron because the antineutrino as an electric dipole interacts under a weak electromagnetic interaction with the charge of the up quark of the stable (dud) quark triad and gives 3d quarks of smaller binding energy than the binding energy of the (dud) triad. Note that in the dud triad we observe both attractive electric and magnetic forces, because the one spinning up quark has a positive charge + 2e/3 and the two down quarks have negative charges. While in the systems ddd of negative charge the magnetic attractions of antiparallel spin are stronger than the electric repulsions, because the peripheral velocity of the spinning quarks is greater than the speed of light. In the same way a dipole photon in the hydrogen atom interacts with the same weak electromagnetic forces with the charge (-e) of the electron and the hydrogen atom becomes an atom of unstable states. ANTINEUTRINO ABSORPTION Before my discovery of quarks in proton and neutron physicists believed that the antineutrino was able to interact with the proton under a fallacious weak interaction including very heavy bosons which violate the two conservation laws of mass and energy: ν- + p = n + e+ However after my discovery of 288 quarks in nucleons the above reaction is written as or ν- + + 4u + 5d = [ (92(dud) + 4u + 8d ] + e+ or ν- + (dud) = 3d + e+ or ν- + u = d + e+ Such a simple interaction based on the natural laws not only invalidates the Electroweak theory but also explains why the energetic antineutrino has a mass equal to the 1.8 MeV/c2 . Because we get d - u = 1.29 MeV/c2 = ν- + e+ = 1.8 + 0.51 Here the antineutrino interacts electromagnetically with the up quark (u) having a positive charge (+2e/3). In the same way a photon as an electric dipole interacts with the charge (-e) in the hydrogen atom and leads to unstable states. Similarly in the antineutrino emission ( β- decay) one observes that the unstable neutron becomes a stable proton because the 3d quarks of neutron become a very stable (dud) quark triad of proton). In the same way the unstable states of the hydrogen atom become a stable hydrogen with a ground state binding energy of 13.6 eV after the emission of a photon. ANTINEUTRINO EMISSION n = p + e- +ν- or + 4u + 8d = [ (93(dud) + 4u + 5d ] + e- + ν- or 3d = (dud) + e- + ν- or d = u + e- + ν- As in the case of the antineutrino absorption here we also see that the antineutrino emission occurs under the same conservation laws of energy and mass. That is d - u = 1.29 MeV = 0.51 + 0.78 However neutrons are stable in nuclei. For example the neutron is stable in the deuteron because the binding energy of deuteron D = 2.2246 MeV is greater than the energy d - u = 1.29 MeV of a free neutron. If a neutron decayed via the process above while in a nucleus, it would produce a proton, but now that proton isn't in isolation like it would be with a lone neutron, it's surrounded by all the other protons in the nucleus. Since the original protons took up all the lower energy states, the lowest possible energy state the new proton could occupy would be a high one, and hence the new state would not be more energetically favorable than just keeping the neutron as it was. Hence, neutrons tend not to decay in stable nuclei having a binding energy greater than the 1.29 MeV. In very massive nuclei, there are loads and loads of neutrons as well, and so it's very possible that you might get a lower energy state by having a neutron decay into a proton. Category:Fundamental physics concepts